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One falls slower in a mine shaft than in free air. This is due to collisions with the walls. How should one model the terminal velocity in the presence of such collisions?

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This looks off-topic... – Malabarba Jan 19 '11 at 13:42
Reminds me of the scene in which Homer Simpson falls into a ravine... – Raskolnikov Jan 19 '11 at 14:04
Miners have told me that if you fall down a deep enough shaft, you're dead before you hit the bottom due to the collisions. – Carl Brannen Jan 26 '11 at 1:49
up vote 2 down vote accepted

Have a look at:

A similar situation is described in:

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the answer would be more helpful if you actually quoted the main result or something – Tobias Kienzler Feb 16 '11 at 12:51

Terminal velocity is terminal velocity, regardless of collisions (Edit: I should clarify this... Terminal velocity is defined as the steady state velocity that you reach when a velocity dependent resistive force exactly cancels the force driving the motion [usually gravity]. Occasionally you'll hit a wall and velocity will be lost deforming the wall (or your body!) These collisions won't reduce the terminal velocity, but they will definitely reduce the average velocity and possibly prevent the falling object from ever even reaching terminal velocity.

As a first pass model, for fun, I'd suggest free fall in gravity, with the usual velocity dependent damping (air resistance) and a stochastic term that reduces the velocity according to a Poisson distribution.

This of course won't change the terminal velocity (which you'd get as the steady state in the collision-less limit), but it would give you a better estimate of the average velocity.

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If the falling body occludes a sizable fraction of the shaft area there can be additional viscus and compressive effects, which would effect the terminal velocity. – dmckee Jan 19 '11 at 2:19
This is quite true, and I don't claim to be an expert on mineshafts -- I kind of imagined it as a pachinko machine with all the pegs on the boundary. :) – wsc Jan 19 '11 at 2:26

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